$h(n) = 6n^{2}-4(f(n))$ $f(x) = -4x^{3}-3x^{2}$ $ f(h(-1)) = {?} $
Answer: First, let's solve for the value of the inner function, $h(-1)$ . Then we'll know what to plug into the outer function. $h(-1) = 6(-1)^{2}-4(f(-1))$ To solve for the value of $h$ , we need to solve for the value of $f(-1)$ $f(-1) = -4(-1)^{3}-3(-1)^{2}$ $f(-1) = 1$ That means $h(-1) = 6(-1)^{2}+(-4)(1)$ $h(-1) = 2$ Now we know that $h(-1) = 2$ . Let's solve for $f(h(-1))$ , which is $f(2)$ $f(2) = -4(2^{3})-3(2^{2})$ $f(2) = -44$